Theorem mpnanrd 697

Description: Eliminate the right side of a negated conjunction in an implication. (Contributed by ML, 17-Oct-2020.)

Hypotheses

Ref	Expression
mpnanrd.1	|- ( ph -> ps )
mpnanrd.2	|- ( ph -> -. ( ps /\ ch ) )

Assertion

Ref	Expression
mpnanrd	|- ( ph -> -. ch )

Proof of Theorem mpnanrd

Step	Hyp	Ref	Expression
1		mpnanrd.1	. 2 |- ( ph -> ps )
2		mpnanrd.2	. . 3 |- ( ph -> -. ( ps /\ ch ) )
3		imnan 694	. . 3 |- ( ( ps -> -. ch ) <-> -. ( ps /\ ch ) )
4	2, 3	sylibr 134	. 2 |- ( ph -> ( ps -> -. ch ) )
5	1, 4	mpd 13	1 |- ( ph -> -. ch )

Syntax hints:   -. wn 3   -> wi 4   /\ wa 104

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  ecase2d  1385